[tex]dane:\\m = 4 \ kg\\T_1 = 20^{o}C\\T_2 = 45^{o}C\\\Delta T = T_2 - T_1 = 45^{o}C - 20^{o}C = 25^{o}C\\c = 130\frac{J}{kg\cdot^{o}C}\\szukane:\\Q = ?\\\\Rozwiazanie\\\\Q = c\cdot m\cdot \Delta T\\\\Q = 130\frac{J}{kg\cdot^{o}C}\cdot4 \ kg\cdot25^{o}C\\\\Q = 13 \ 000 \ J = 13 \ kJ[/tex]
